Calculate required n for the confidence interval width to be doubled

[Harry_the_cat]

Continued from Post#20

Simplifying the LHS gives:
\(\displaystyle w = 2z\sqrt{\frac{p(1-p)}{n}}\)
\(\displaystyle w = 2z*\frac{\sqrt{p(1-p)}}{\sqrt{n}}\)

When n=100:
\(\displaystyle w = 2z*\frac{\sqrt{p(1-p)}}{\sqrt{100}}\)
\(\displaystyle w = 2z*\frac{\sqrt{p(1-p)}}{10}\)

Now, if the new \(\displaystyle w\) is to be double this, we need the denominator to be 5 not 10, ie the new \(\displaystyle n\) must be 25.